What are the values of $a$ that satisfy the equation $\frac{ax^2+ax-1}{2(x+\frac{5}{4})^2+\frac{47}{8}} < 0$?

Question

I got this question in a test today, I was stumped on how to do it. I am right to say that since $2(x+\frac{5}{4})^2+\frac{47}{8}$ is always positive $ax^2+ax-1$ is always negative. So, $$ax^2+ax-1<0$$$$Discriminant < 0$$$$(a)^2-4(a)(-1)<0$$$$a(a+4)<0$$$$-4<a<0$$ or is it $$ax^2+ax-1<2x^2+5x+9$$$$(a-2)x^2+(a-5)x-10<0$$$$Discriminant < 0$$$$(a-5)^2-4(a-2)(-10)<0$$

Answer 1

Hint: You want $a < 0$, and $a^2 - 4a(-1) < 0$